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Lady bird [3.3K]
3 years ago
13

Suzy brought 50 apples from the orchard she brought 5 times as many red apple than green apples how many green apples did she bu

y
Mathematics
1 answer:
sveta [45]3 years ago
8 0
You have to multiply 50×5 which is 250
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Solve this equation
Novosadov [1.4K]
B. 9^2

explanation: 3^4 is equivalent to 3*3*3*3, which is 81
9^2 (9*9) also equals 81
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2 years ago
How much do you pay tax per each 1 dollar
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Depends on your state. Each state is different. I'll try to post the complete list.

Hope this is what you are looking for!

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3 years ago
Read 2 more answers
One hundred items are simultaneously put on a life test. Suppose the lifetimes
romanna [79]

Answer:

a) \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b) \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( i-1 )st and the ith failures. Then, the T_{i} are independent with \mathrm{T}_{\mathrm{i}} being exponential with rate \frac{(101-i)}{200} .

Therefore,

a) E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]

=\sum_{i=1}^{5} \frac{200}{101-i}

\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b)

The variance is given by, \mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]

\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

7 0
3 years ago
How to put this in $799.73 in numeric please help thank you
AlekseyPX
The answer is 800$...
6 0
3 years ago
Drag the tiles to list the sides of △MNO from shortest to longest.
sweet [91]

The smaller the angle subtended by a side, the smaller the length of the

side.

The correct responses are;

Question 1: The list of sides from shortest to longest are;

  • MO/Shortest MO/Medium and MO/Longest

a) <u>Friday</u>

b) <u>70 minutes</u>

c) <u>40%</u>

d) Yes<u>,</u> <u>the sum of the </u><u>mean</u><u> number of </u><u>minutes spent</u><u> on </u><u>aerobic</u><u> training and the mean number of minutes spent on </u><u>strength</u><u> training is equal to the mean </u><u>total</u><u> number of minutes spent </u><u>training.</u>

From the given diagram, we have, the measure of the third angle, ∠O, is

found as follows;

∠O = 180° - 54° - 61° = 65°

Therefore, ∠O = The largest angle

We get;

The longest side is opposite the largest angle, which gives;

The shortest side is the side opposite ∠N (54°)= \frac{}{MO}

The next shortest side is the side opposite ∠M(61°) = \frac{}{NO}

The longest side is the side opposite ∠O(65°) = \frac{}{MN}

a) The time spent training on Tuesday = 60 + 10 = 70 minutes

The time spent training on Thursday = 50 + 30 = 80 minutes

The time spent training on Friday = 45 + 40 = 85 minutes

Therefore, the day the athlete spent the longest total amount of time training is on <u>Friday</u>

b) The time spent training on Monday = 10 + 20 = 30 minutes

The time spent training on Wednesday = 20 + 15 = 35 minutes

Therefore, we get;

30, 35, 70, 80, and 85

The median total number of minutes the athlete spent training each day = <u>70 minutes</u>

<u />

c) The time spent strength training = 20 + 10 + 15 + 30 + 45 = 120

The total number of minutes the athlete spent training = 70 + 80 + 85 + 30 + 35 = 300

The  percentage spent on strength training = \frac{120}{300} × 100 = \frac{40}%

d) The mean number of minutes spent on strength training is found as follows;

Mean_{strength} =\frac{120}{5} =24

The mean number of minutes spent on aerobic training is found as follows;

Mean_{aerobic} =\frac{10+60+20+50+40}{5} =36

Mean_{strength} +Mean_{aerobic} =24+36=60

The mean total number of minutes spent training, Mean_{total} = \frac{300}{5} = 60

Therefore;

  • Mean_{strength}+Mean_{aerobic} = Mean_{total} \\

Learn more here:

brainly.com/question/2962546

4 0
2 years ago
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